mirror of
https://github.com/donnemartin/interactive-coding-challenges.git
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159 lines
3.9 KiB
Plaintext
159 lines
3.9 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Problem: Find the kth to last element of a linked list\n",
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"\n",
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"* [Clarifying Questions](#Clarifying-Questions)\n",
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"* [Test Cases](#Test-Cases)\n",
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"* [Algorithm](#Algorithm)\n",
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"* [Code](#Code)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Clarifying Questions\n",
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"\n",
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"* Is k an integer?\n",
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" * Yes\n",
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"* If k = 0, does this return the last element?\n",
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" * Yes\n",
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"* What happens if k is greater than or equal to the length of the linked list?\n",
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" * Return None\n",
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"* Can you use additional data structures?\n",
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" * No"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Test Cases\n",
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"\n",
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"* Empty list\n",
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"* k is not an integer\n",
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"* k is >= the length of the linked list\n",
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"* One element, k = 0\n",
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"* General case with many elements, k < length of linked list"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Algorithm\n",
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"\n",
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"* Check for edge cases above, returning None for errors\n",
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"* Setup two pointers, current and previous\n",
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"* Give current a headstart, incrementing it once for k = 1, twice for k = 2, etc\n",
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"* Increment both pointers until current reaches the end\n",
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"* Return the value of previous\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n)\n",
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"* Space: In-place"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": true
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},
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"outputs": [],
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"source": [
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"%run linked_list.py"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"class MyLinkedList(LinkedList):\n",
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" def kth_to_last_elem(self, k):\n",
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" if self.head is None:\n",
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" return\n",
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" if type(k) != int:\n",
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" raise ValueError('')\n",
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" if k >= len(self):\n",
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" return\n",
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" curr = self.head\n",
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" prev = self.head\n",
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" counter = 0\n",
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" while counter < k:\n",
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" curr = curr.next\n",
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" counter += 1\n",
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" if curr is None:\n",
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" return\n",
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" while curr.next is not None:\n",
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" curr = curr.next\n",
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" prev = prev.next\n",
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" return prev.data"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"# Empty list\n",
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"linked_list = MyLinkedList(None)\n",
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"print(linked_list.kth_to_last_elem(0))\n",
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"# k is not an integer\n",
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"print(linked_list.kth_to_last_elem('a'))\n",
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"# k is >= the length of the linked list\n",
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"print(linked_list.kth_to_last_elem(100))\n",
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"# One element, k = 0\n",
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"head = Node(2)\n",
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"linked_list = MyLinkedList(head)\n",
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"print(linked_list.kth_to_last_elem(0))\n",
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"# General case with many elements, k < length of linked list\n",
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"linked_list.insert_to_front(1)\n",
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"linked_list.insert_to_front(3)\n",
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"linked_list.insert_to_front(5)\n",
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"linked_list.insert_to_front(7)\n",
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"print(linked_list.kth_to_last_elem(2))"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 2",
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"language": "python",
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"name": "python2"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 2
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython2",
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"version": "2.7.9"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 0
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